
MHB Master
#21
February 10th, 2020,
01:41
Thread Author
Originally Posted by
Klaas van Aarsen
I didn't check them all, but indeed it can't be right that $2$ is in this list twice, nor that it maps to two different elements.
Why is it in the list twice?
We start from the rightmost cycle, right? Then we have the element $2$ and see that this goes to $4$ and $4$ goes to $9$, so $2$ goes to $9$.
At the second cycle we have again the element $2$. When we reach this cycle do we check again the element $2$ ?

February 10th, 2020 01:41
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MHB Seeker
#22
February 10th, 2020,
04:58
Originally Posted by
mathmari
We start from the rightmost cycle, right? Then we have the element $2$ and see that this goes to $4$ and $4$ goes to $9$, so $2$ goes to $9$.
At the second cycle we have again the element $2$. When we reach this cycle do we check again the element $2$ ?
No.
When we consider the image of an element such as $2$, we must evaluate all cycles from right to left. We are not supposed to start in the middle.

MHB Master
#23
February 10th, 2020,
10:45
Thread Author
Ok !!
Now I think I got the right results :
$\pi_4=\pi_1\circ\pi_2=(14\ 13) \ (6\ 3\ 2\ 9\ 1)\ (12\ 8\ 7\ 4\ 5)$
$\pi_5=\pi_2\circ\pi_3=(1\ 4) \ (2\ 8\ 11\ 12\ 6\ 7)\ (13\ 14)$
$\pi_6=\pi_2\circ\pi_1=(3 \ 4 \ 9 \ 12\ 1) \ (2\ 5\ 8\ 6\ 7)\ (13 \ 14)$
For the signum we have to write each permutation as a product of cycles of two elements, right? How can we do that?

MHB Seeker
#24
February 10th, 2020,
13:02
Originally Posted by
mathmari
For the signum we have to write each permutation as a product of cycles of two elements, right? How can we do that?
We can write $(a\,b\,c\,d)$ for instance as either $(a\,d)(a\,c)(a\,b)$ or $(a\,b)(b\,c)(c\,d)$.
Note that more generally a cycle with an even number of elements can be written as an odd number of 2cycles (signum 1).
And a cycle with an odd number of elements can be written as an even number of 2cycles (signum +1).

MHB Master
#25
February 20th, 2020,
02:13
Thread Author
I am looking aagain at the previous question, at the calculation of $\pi_4$. Do we get the same results if we use the initially given $\pi_1$ and $\pi_2$ and the same if we use the cycle decomposition of them?
I get the following from the cycle decomposition:
$\pi_1=(8 \ 6 \ 12)\ (1 \ 9 \ 4 \ 2 \ 3 )$
$\pi_2=(7 \ 8 \ 5 \ 4 \ 2)\ (6 \ 12 \ 1 ) \ (14 \ 13)$
$\pi_4=\pi_1\circ\pi_2=(8 \ 6 \ 12)\ (1 \ 9 \ 4 \ 2 \ 3 ) \ (7 \ 8 \ 5 \ 4 \ 2)\ (6 \ 12 \ 1 ) \ (14 \ 13)$
$14\rightarrow 13$
$13\rightarrow 14$
$6\rightarrow 12\rightarrow 8$
$12\rightarrow 1\rightarrow 9$
$1\rightarrow 6\rightarrow 12$
$7\rightarrow 8\rightarrow 6$
$8\rightarrow 5$
$5\rightarrow 4\rightarrow 2$
$4\rightarrow 2\rightarrow 3$
$2\rightarrow 7$
$9\rightarrow 4$
$3\rightarrow 1$
So $\pi_4=(14 \ 13) \ (6 \ 8 \ 5 \ 2 \ 7) \ (12 \ 9 \ 4 \ 3 \ 1) $.
And from the initially given forms:
$\pi_1=(1 \ 2\ 4 \ 9)\ (1\ 3)\ (6 \ 8 \ 12)$
$\pi_2=(2 \ 4\ 5 \ 8 \ 7)\ ( 1 \ 12 \ 6) \ (13 \ 14)$
$\pi_4=\pi_1\circ\pi_2=(1 \ 2\ 4 \ 9)\ (1\ 3)\ (6 \ 8 \ 12) \ (2 \ 4\ 5 \ 8 \ 7)\ ( 1 \ 12 \ 6) \ (13 \ 14)$
$13 \rightarrow 14$
$14\rightarrow 13$
$1\rightarrow 12 \rightarrow 6$
$12\rightarrow 6\rightarrow 8$
$6\rightarrow 1 \rightarrow 3$
$2\rightarrow 4\rightarrow 9$
$4\rightarrow 5$
$5\rightarrow 8\rightarrow 12$
$8\rightarrow 7$
$7\rightarrow 2\rightarrow 4$
$3\rightarrow 1\rightarrow 2$
$9\rightarrow 1$
So $\pi_4=(13 \ 14) \ (1 \ 6 \ 3 \ 2 \ 9 ) \ (12 \ 8 \ 7 \ 4 \ 5)$.
Have I done something wrong or is it correct that we get different results?

MHB Seeker
#26
February 20th, 2020,
05:37
The original $\pi_2$ was already disjoint and contained (1 12 6). But it's not like that in both your cases, is it?

MHB Master
#27
February 20th, 2020,
19:24
Thread Author
Originally Posted by
Klaas van Aarsen
The original $\pi_2$ was already disjoint and contained (1 12 6). But it's not like that in both your cases, is it?
What do you mean? I got stuck right now.

MHB Seeker
#28
February 20th, 2020,
20:18
Originally Posted by
mathmari
I get the following from the cycle decomposition:
$\pi_1=(8 \ 6 \ 12)\ (1 \ 9 \ 4 \ 2 \ 3 )$
$\pi_2=(7 \ 8 \ 5 \ 4 \ 2)\ (6 \ 12 \ 1 ) \ (14 \ 13)$
You wrote (6 12 1) here.
Originally Posted by
mathmari
And from the initially given forms:
$\pi_1=(1 \ 2\ 4 \ 9)\ (1\ 3)\ (6 \ 8 \ 12)$
$\pi_2=(2 \ 4\ 5 \ 8 \ 7)\ ( 1 \ 12 \ 6) \ (13 \ 1
4)$
And you wrote (1 12 6) here.
Shouldn't they be the same?
Did you perhaps accidentally write $\pi_2^{1}$ instead of the decomposition?

MHB Master
#29
February 21st, 2020,
01:43
Thread Author
Originally Posted by
Klaas van Aarsen
You wrote (6 12 1) here.
And you wrote (1 12 6) here.
Shouldn't they be the same?
Did you perhaps accidentally write $\pi_2^{1}$ instead of the decomposition?
Oh yes, you are right! I used theinverse accidentaly.
So we get the same result using the initial permutation and the cycle decomposition of it.